im trying to provide power to my PCB using a 9V battery. however, the 10 pins on the left use 3.3V and the 10 pins on the right use 5V. I know i should use a voltage regulator to do so, but im afraid that my circuits basics are not that good.

could someone plz explain to me how the voltage regulator circuit should be connected.

As far as I can see there is only 5 Volts used.
The 3.3 Volts is made on the board itself from the 5 Volts input.
The 5 Volts must be connected to pins 2,3 and 9.
The Ground connection to pins 12,13,15 and 26.

You can use the 5 volts regulator.
I do not know how long a 9 volts battery will last.
There is stated in the PDF the 3.3 volts may deliver 200 mA, wich is quite a lot for a 9 volts battery.

As bertus says all you need is a 5 Volt input, the 3.3 Volt is an output from the board.

One point I would make is that if you are using a 9 Volt battery I would not use an LM7805 regulator. The LM7805 has a dropout voltage of 2 Volt which means that it needs 2 volts dropped across the regulator for it to work. Once your on load battery voltage drops to 7 volt the regulator will turn off. This will give very poor battery life.

Instead I would go for a low dropout regulator such as an LM2931-5 or LM2937-5 which have a dropout voltage of about 0.5 Volt so the regulator will keep working till the battery voltage falls to about 5.5 Volt.

It's a 20-minute demo. Four, 2300 mAh AA NiMH batteries would give you a pretty steady voltage of 5.2'ish to 4.8 at 2.3A for that period -- maybe even better, if your current drain isn't much. Or, you might even consider four, 1000 mAh AAA batteries.